A certain car traveled twice as many miles from Town A to Town B as it

Question

A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?

13
13.5
14
14.5
15

Can some one help me solve this problem using weighted averages? Why cannot we use gallons as weights? Thanks.

Source: GMAT club problem

A. 13
B. 13.5
C. 14
D. 14.5
E. 15

Bunuel · Accepted Answer

Average miles per gallon equals to total miles covered/total gallons used (miles/gallons), so if the distance from A to B is 2x miles and from B to C is x miles then we'll have: average \ miles \ per \ gallon=\frac{total \ miles \ covered}{total \ gallons \ used}=\frac{2x+x}{\frac{2x}{12}+\frac{x}{18}}=\frac{3x}{\frac{2x}{9}}=13.5 . Answer: B. Now, weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2} and as we are asked to determine average miles per gallon then exactly miles per gallon should be the weights and gallons used for 1st and 2nd parts of the trip should be the values, so in this case the weighted average formula will give the same expression as the one above. Hope it's clear. P.S. Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/ Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/ No posting of PS/DS questions is allowed in the main Math forum.

Bunuel · Answer

Average  miles per gallon  equals to  total miles covered/total gallons used  (miles/gallons), so if the distance from A to B is  2x  miles and from B to C is  x  miles then we'll have:
 average \ miles \ per \ gallon=\frac{total \ miles \ covered}{total \ gallons \ used}=\frac{2x+x}{\frac{2x}{12}+\frac{x}{18}}=\frac{3x}{\frac{2x}{9}}=13.5 .

Answer: B.

P.S. As you can see weighted average concept is exactly what we used here.

Hope it's clear.

KarishmaB · Answer

Note: Whenever you are confused what the weights should be (e.g. here should the weights be the distance traveled should they be gallons of fuel used...) look at the units.

Average required is  \frac{miles}{gallon} . So you are trying to find the weighted average of two quantities whose units must be  \frac{miles}{gallon} .

C_{avg} = \frac{C_1*W_1 + C_2*W_2}{{W_1 + W_2}}

C_{avg}, C_1, C_2 - \frac{miles}{gallon}

So  W_1  and  W_2  should be in gallon to get:

\frac{miles}{gallon} = (\frac{miles}{gallon}*gallon + \frac{miles}{gallon}*gallon)/(gallon + gallon)

Food for thought: If I sold 10 apples at a profit of 10% and 15 oranges at a profit of 20%, what was my overall profit%?

shankar245 · Answer

Karishma,

Why is this not working
?

12*(2/3) + 18 *(1/3)
/1

14??

KarishmaB · Answer

We need to find the average miles per gallon.

Average miles/gallon = Total miles/Total gallons

What is the distance travelled (i.e. miles)? 
"traveled twice as many miles from Town A to Town B as it did from Town B to Town C"
We know that the ratio of the two distances must be 2:1 or we can say the distances must be 2d and d.
What is the total gallons used?
Fuel used to go from A to B = 2d/12
Fuel used to go from B to C = d/18

So Average miles/gallon = (2d + d)/(2d/12 + d/18)

Go back to my post above. It explains that you have to be mindful of how to use weighted averages. You need to know what the weights are going to be. Here, weights have to be the amount of fuel used i.e. in gallons because you are looking for average miles per gallon.

If we change the question and make it:
A certain car used twice as many gallons from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?

Now you can use 12*(2/3) + 18 *(1/3)/1

I hope it makes sense now.

MSoS · Answer

Lets pick numbers:

Distance A to B: 36 miles (2x) --> 3 gallons
Distance B to C: 18 miles (x) --> 1 gallon

Average:  \frac{(36+18)}{4}  = 13.5

Hence, B

KarishmaB · Answer

If the concept you do not understand is why (12*2 + 18*1)/3 doesn't work, here you go:

'Distances traveled' (i.e. ratio of 2:1) cannot be the weights here to find the average mileage. The weights have to be 'number of gallons'.

If we change the question and make it:
A certain car used  twice as many gallons  from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?

Now you can use (12*2 + 18 *1)/3

Why?

Whenever you are confused what the weights should be (e.g. here should the weights be the distance traveled or should they be gallons of fuel used...), look at the units.

Average required is  \frac{miles}{gallon} . So you are trying to find the weighted average of two quantities whose units must be  \frac{miles}{gallon} .

C_{avg} = \frac{C_1*W_1 + C_2*W_2}{{W_1 + W_2}}

C_{avg}, C_1, C_2 - \frac{miles}{gallon}

So  W_1  and  W_2  should be in gallon to get:

\frac{miles}{gallon} = (\frac{miles}{gallon}*gallon + \frac{miles}{gallon}*gallon)/(gallon + gallon)

Only if weights are in gallons, do we get 'Total miles' in the numerator and 'Total gallons' in the denominator.

We know that Average miles/gallon = Total miles/Total gallons

Takeaway: The weights have to be the denominator units of the average.

So what do we do in this question?

What is the distance travelled (i.e.total miles)? 
"traveled twice as many miles from Town A to Town B as it did from Town B to Town C"
We know that the ratio of the two distances must be 2:1 or we can say the distances must be 2d and d. Total distance must be (2d + d)

What is the total gallons used?
Fuel used to go from A to B = 2d/12
Fuel used to go from B to C = d/18

So Average miles/gallon = (2d + d)/(2d/12 + d/18)

You need to know what the weights are going to be. Here, weights have to be the amount of fuel used i.e. in gallons because you are looking for average miles per gallon.

PareshGmat · Answer

Let distance A to B = 2, then distance B to C = 1

Overall Average  = \frac{Total distance}{Total consumption} = \frac{2+1}{\frac{2}{12} + \frac{1}{18}} = \frac{3}{4} * 18 = 13.5

Answer = B

ScottTargetTestPrep · Answer

We can use the following formula:

Average = (total distance)/(total gallons)

We can let the distance from Town A to Town B = 2d and the distance from Town B to Town C = d. Thus, the total gallons of fuel consumed from Town A to Town B = 2d/12, and the total gallons of fuel consumed from Town B to Town C = d/18; thus:

Average = 3d/(2d/12 + d/18)

Average = 3d/(6d/36 + 2d/36)

Average = 3d/(8d/36) = (36 x 3d)/8d = (9 x 3)/2 = 13.5

Answer: B

EMPOWERgmatRichC · Answer

Hi All,

We're told that a certain car traveled TWICE as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. We're asked for the average MILES PER GALLON that the car achieved on its trip from Town A through Town B to Town C. This question can be solved in a couple of different ways, including by TESTing VALUES.

To start, we should look to pick distances that work well with the given miles/gallon data that we have (re: 12 miles/gallon and 18 miles/gallon). Choosing 36 miles for the first part of the trip and 18 miles for the second part of the trip will make the math fairly easy.

Town A to Town B: 36 miles traveled at 12 miles/gallon --> 36/12 = 3 gallons used to travel 36 miles

Town B to Town C: 18 miles traveled at 18 miles/gallon --> 18/18 = 1 gallon used to travel 18 miles

Total miles traveled = 36 + 18 = 54 miles 
Total gallons used = 3 + 1 = 4 gallons 
Average miles/gallon = 54/4 = 13.5

Final Answer:  B

GMAT assassins aren't born, they're made, 
Rich

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